A contradiction to do with continuity? (involves chain rule) Suppose for each $t$, $S(t) \subset \mathbb{R}^n$ is a domain (hypersurface). We have a diffeomorphism $D^0_t:S(0) \to S(t)$ for each $t$ such that it solves the ODE
$$\frac{d}{dt}D^0_t(\cdot) = V(D^0_t(\cdot),t)$$
$$D^0_0(\cdot) = \text{Identity}(\cdot)$$
where $V$ is discontinuous wrt. $t$. We also know that $D^0_t$ is $C^2$ in space.
Define the map
$F_{-t}:H^1(S(t)) \to H^1(S(0))$ by
$$(F_{-t} u)(x) = u(D^0_t(x)),$$
and define $F_t$ similarly. So $F_{-t}$ takes a function defined on $S(t)$ and turns into a function defined on $S(0)$, and $F_t$ does the opposite.
Suppose that for each $t$, $u(t):S(t) \to \mathbb{R}$ is a function defined on $S(t)$. Then
$$(F_{-t}u(t))(\cdot) = u(t, D^0_t(\cdot))$$
and so 
$$\frac{d}{dt}(F_{-t}u(t))(\cdot) = \frac{d}{dt}u(t, D^0_t(\cdot)) = u_t|_{(t, D^0_t(\cdot))} + \nabla u|_{(t, D^0_t(\cdot))}\cdot V(D_t^0(\cdot),t).\tag{1}$$
Now suppose that $v \in C^1([0,T];H^1(S_0))$, so
$$v(t):S(0) \to \mathbb{R}$$
for all $t$. Consider $u(t) := F_t v(t)$ for all $t$. Clearly $u(t):S(t) \to \mathbb{R}$. In this case then,
$$\frac{d}{dt}(F_{-t}u(t))(\cdot) = \frac{d}{dt}(F_{-t}F_tv(t))(\cdot) = \frac{d}{dt}(v(t))(\cdot)$$
which is continuous wrt $t$ by definition. But this expression is also equal to (1), the right hand side of which is NOT continuous wrt. $t$ because of the $V$ present which is not continuous. So what is going on? Isn't this contradictory?
 A: The reason this is not a contradiction is that the discontinuity of $V$ does NOT imply the discontinuity of expressions involving $V$.
In this case you need to be careful about exactly what you're taking the domain of $u$ to be - the only natural choice that makes $u$ distinct from $v$ seems to be the subspace of $\mathbb R^n$ that is swept out by the hypersurfaces $S(t)$. In this case $u$ has no $t$ dependence (since the coordinate in space determines which $S(t)$ you are on), and $(1)$ should simply read
$$
v_t(\cdot) = \frac{d}{dt}(F_{-t}u)(\cdot) = \nabla u|_{D^0_t(\cdot)}\cdot V(D_t^0(\cdot)).
$$
The reason the RHS is not discontinuous is that $\nabla u$ is inseperably linked to the velocity of the flow by its definition - when $V$ is small, the spacing between $S(t)$ is small, and thus $u$ will change faster when moving in the direction of increasing $t$. This means any discontinuity in $V$ is transferred into a discontinuity in $\nabla u$, exactly as described by $v_t = \nabla u \cdot V$.
